JEE Main 2026 — 23 January, Evening Shift — Question 21

Given: $(x^2-4)y' - 2xy + 2x(4-x^2)^2 = 0$, $x>2$. Rewrite as: $(x^2-4)y' - 2xy = -2x(4-x^2)^2$. Divide by $(x^2-4)^2$: $\frac{y'}{x^2-4} - \frac{2xy}{(x^2-4)^2} = -2x\frac{(4-x^2)^2}{(x^2-4)^2}$. Note that $\frac{d}{dx}\left(\frac{y}{x^2-4}\right) = \frac{y'}{x^2-4} - \frac{2xy}{(x^2-4)^2}$. Also, $\frac{(4-x^2)^2}{(x^2-4)^2} = 1$. So the equation becomes: $\frac{d}{dx}\left(\frac{y}{x^2-4}\right) = -2x$. Integrate: $\frac{y}{x^2-4} = -x^2 + C$. Thus, $y = (x^2-4)(-x^2 + C)$. Given $y(3)=15$: $15 = (9-4)(-9+C) = 5(C-9)$ $\Rightarrow C-9=3 \Rightarrow C=12$. So $y = (x^2-4)(-x^2+12) = -(x^2-4)(x^2-12)$. For local maximum, $y'=0$: $y' = -[2x(x^2-12) + (x^2-4)(2x)] = -2x(2x^2-16)=0$. Since $x>2$, $x=0$ is not allowed, so $2x^2-16=0 \Rightarrow x^2=8 \Rightarrow x=2\sqrt{2}$. Then $y_{\text{max}} = (8-4)(-8+12) = 4 \times 4 = 16$. Hence, the local maximum value is 16.